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This statement is in a book I am reading

Let $0<\alpha<1$. Then the change of variables $u=k+(s-k)z$ tells us that

$\frac{1}{\Gamma(\alpha)}\int_k^s(u-k)^{-\alpha}(s-u)^{\alpha-1}du=\Gamma(1-\alpha).$

But why is this true? I get that

$\int_k^s(u-k)^{-\alpha}(s-u)^{\alpha-1}du=\int_0^1(s-k)^{-\alpha}z^{-\alpha}(s-k-(s-k)z)^{\alpha-1}dz(s-k)$

$=\int_0^1z^{-\alpha}(1-z)^{\alpha-1}dz$.

But then I still have to show that

$$ \frac{1}{\Gamma(\alpha)}\int_0^1z^{-\alpha}(1-z)^{\alpha-1}dz=\Gamma(1-\alpha).$$

But I do not know how to evaluate the integral $\int_0^1z^{-\alpha}(1-z)^{\alpha-1}dz$, or how to show this.

Can you please help?, do you have any tips?

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    your last integral is a standard representation of euler's beta function – 2017-01-27
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    to prove this write $\Gamma(x)\Gamma(y)=4\int_0^{\infty}dq\int_0^{\infty}dl l^{2x}q^{2y}e^{-q^2-l^2}$. Now change to polar coordinates – 2017-01-27

1 Answers 1

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We know that the definition of the Beta function is : $$B (l,m) =\int_{0 }^{1} x^{l-1}(1-x)^{m-1} \mathrm {d}x $$ and the relation between the beta function and the gamma function is: $$B (l,m) =\frac {\Gamma(l)\Gamma (m)}{\Gamma (l+m)}, l>0, m>0$$

Can you take it from here? Hope it helps.